The function \( f \) is defined by \( f(x)=x^{3}+2 \sin (2 x+2) \). If the function \( g \) is defined to be \( g(x)=\int_{0}^{x} f(t) d t \), what is the average rate of change of \( g \) on the interval \( [-3,1] \) ? You may use a calculator and round your answer to the nearest thousandth.

To find the average rate of change of the function \( g \) on the interval \([-3, 1]\), we use the formula for the average rate of change: \[ \text{Average Rate of Change} = \frac{g(b) - g(a)}{b - a} \] where \( a = -3 \) and \( b = 1 \). First, we need to calculate \( g(1) \) and \( g(-3) \). Given that \( g(x) = \int_0^x f(t) \, dt \), we have: \[ g(1) = \int_0^1 f(t) \, dt \quad \text{and} \quad g(-3) = \int_0^{-3} f(t) \, dt \] Since \( g(-3) \) from \( 0 \) to \(-3\) can be reversed, we have: \[ g(-3) = -\int_{-3}^0 f(t) \, dt \] Now we need to compute \( g(1) \) and \( g(-3) \). ### Step 1: Calculate \( g(1) \) \[ g(1) = \int_0^1 (t^3 + 2 \sin(2t + 2)) \, dt \] This integral can be split into two parts: \[ g(1) = \int_0^1 t^3 \, dt + 2 \int_0^1 \sin(2t + 2) \, dt \] Calculating the first integral: \[ \int_0^1 t^3 \, dt = \left[\frac{t^4}{4}\right]_0^1 = \frac{1}{4} \] Now for the second integral, using substitution \( u = 2t + 2 \) gives \( du = 2dt \) or \( dt = \frac{du}{2} \). When \( t = 0 \), \( u = 2 \) and when \( t = 1 \), \( u = 4 \). \[ \int_0^1 \sin(2t + 2) \, dt = \frac{1}{2} \int_2^4 \sin(u) \, du = \frac{1}{2} [-\cos(u)]_2^4 = \frac{1}{2} \left[-\cos(4) + \cos(2)\right] \] Plugging in approximate values (using a calculator): \[ \cos(4) \approx -0.6536, \quad \cos(2) \approx -0.4161 \] So, \[ -\cos(4) + \cos(2) \approx 0.6536 - 0.4161 \approx 0.2375 \] Therefore, \[ g(1) = \frac{1}{4} + 2 \cdot \frac{0.2375}{2} = \frac{1}{4} + 0.2375 \approx 0.25 + 0.2375 \approx 0.4875 \] ### Step 2: Calculate \( g(-3) \) We calculate: \[ g(-3) = -\int_{-3}^0 f(t) \, dt = -\left( \int_{-3}^0 (t^3 + 2 \sin(2t + 2)) \, dt \right) \] Using the same method: 1. Calculate \( \int_{-3}^0 t^3 \, dt = \left[\frac{t^4}{4}\right]_{-3}^0 = 0 - \frac{(-3)^4}{4} = -\frac{81}{4} = -20.25 \) 2. Calculate \( \int_{-3}^0 2\sin(2t + 2) \, dt \) using substitution \( u = 2t + 2 \): - When \( t = -3 \), \( u = -4 \) and when \( t = 0 \), \( u = 2 \), \[ \int_{-3}^0 2\sin(2t + 2) \, dt = \int_{-4}^2 \sin(u) \, du = [-\cos(u)]_{-4}^2 = -\cos(2) + \cos(-4) \] Which simplifies to (as cosine is even): \[ -\cos(